Python 给定的查询字符串查找一组具有相同开头的字符串

Question

编辑：我很欣赏所有的答案，但谁能告诉我为什么我的解决方案不起作用？我想尝试在没有 .startswith(( 的情况下执行此操作，谢谢！

我正在尝试完成此练习：

实施自动完成系统。也就是说，给定一个查询字符串和一组所有可能的查询字符串， 返回集合中以 s 为前缀的所有字符串。 例如，给定查询字符串 de 和字符串集 [dog， deer， deal]，返回 [deer， deal]。 提示：尝试将字典预处理为更高效的数据结构，以加快查询速度。

但我得到一个空列表。我可能做错了什么？我以为这会给我[鹿，交易]

def autocomplete(string,set):
string_letters = []
letter_counter = 0
list_to_return = []
for letter in string:
string_letters.append(letter)
for words in set:
for letter in words:
if letter_counter == len(string):
list_to_return.append(words)
if letter == string_letters[letter_counter]:
letter_counter += 1
else:
break
return list_to_return
print(autocomplete("de", ["dog","deer","deal"]))

输出：

[]

编辑：我很欣赏所有的答案，但谁能告诉我为什么我的解决方案不起作用？我想尝试在没有 .startswith(( 的情况下执行此操作，谢谢！

Answer 1

以下是我将如何完成您要做的事情：

import re
strings = ['dog', 'deer', 'deal']
search = 'de'
pattern = re.compile('^' + search)
[x for x in strings if pattern.match(x)]

结果：['deer', 'deal']

但是，在大多数情况下，对于此类用例，您可能希望忽略搜索字符串和搜索字段的大小写。

import re
strings = ['dog', 'Deer', 'deal']
search = 'De'
pattern = re.compile('^' + search, re.IGNORECASE)
[x for x in strings if pattern.match(x)]

结果：['Deer', 'deal']

要回答代码不起作用的部分原因，向代码添加一些冗长的内容会有所帮助：

def autocomplete(string,set):
string_letters = []
letter_counter = 0
list_to_return = []
for letter in string:
string_letters.append(letter)
for word in set:
print(word)
for letter in word:
print(letter, letter_counter, len(string))
if letter_counter == len(string):
list_to_return.append(word)
if letter == string_letters[letter_counter]:
letter_counter += 1
else:
print('hit break')
break
return list_to_return
print(autocomplete("de", ["dog","deer","deal"]))

输出：

dog
('d', 0, 2)
('o', 1, 2)
hit break
deer
('d', 1, 2)
hit break
deal
('d', 1, 2)
hit break
[]

正如您在狗的输出中看到的那样，"d 匹配但 o 没有"，这使得letter_counter 1，然后在鹿 'd ！= 'e' 所以它断裂......这种情况一遍又一遍地持续下去。有趣的是，由于这种行为，设置"ddeer"实际上会匹配。要解决此问题，您需要重置 for 循环中的letter_counter，并具有额外的断点以防止过度修订索引。

def autocomplete(string,set):
string_letters = []
list_to_return = []
for letter in string:
string_letters.append(letter)
for word in set:
# Reset letter_counter as it is only relevant to this word.
letter_counter = 0
print(word)
for letter in word:
print(letter, letter_counter, len(string))
if letter == string_letters[letter_counter]:
letter_counter += 1
else:
# We did not match break early
break
if letter_counter == len(string):
# We matched for all letters append and break.
list_to_return.append(word)
break
return list_to_return
print(autocomplete("de", ["dog","deer","deal"]))

Answer 2

我注意到了提示，但它没有说明为要求，所以：

def autocomplete(string,set):
return [s for s in set if s.startswith(string)]
print(autocomplete("de", ["dog","deer","deal"]))

str.startswith(n)将返回一个布尔值，True如果 str 以n开头，否则False。

Answer 3

您可以使用startswith字符串函数并避免所有这些计数器，如下所示：

def autocomplete(string, set):
list_to_return = []
for word in set:
if word.startswith(string):
list_to_return.append(word)
return list_to_return
print(autocomplete("de", ["dog","deer","deal"]))

Answer 4

简化。

def autocomplete(string, set):
back = []
for elem in set:
if elem.startswith(string[0]):
back.append(elem)
return back
print(autocomplete("de", ["dog","deer","deal","not","this","one","dasd"]))